＆＃34;选择＆＃34;存储过程中的语句不起作用

Question

我尝试使用内连接实现一个存储过程来显示来自不同表的结果，但我的问题是select语句没有返回任何结果，但是它将一些值打印为消息。

alter proc EmployeeReport(@empid int)
as
begin
declare @inTime time(0)
declare @outTime time(0)
declare @fromDate date
declare @toDate date

set @inTime = (select CAST(InTime as time(0)) from Timelog where EmployeeId=@empid)
set @outTime = (select CAST(OutTime as time(0)) from Timelog where EmployeeId = @empid)
set @fromDate = (select cast (InTime as date) from Timelog where EmployeeId= @empid)
set @toDate = (select cast (outTime as date) from Timelog where EmployeeId= @empid)

select @fromDate as FromDate
      ,@toDate as ToDate
      ,c.Name as Client
      ,p.Name as Project
      ,@inTime as InTime
      ,@outTime as OutTime
      ,t.TotalTime
from Timelog t
    inner join Employee e
        on e.id = t.EmployeeId
    inner join Project p
        on p.Id = t.EmployeeProjectId
    inner join Client c
        on c.Id = p.ClientId
where t.EmployeeId = @empid

 print @inTime
 print @outTime
 print @fromDate
 print @toDate
 end

我正在附加我得到的输出文件，请帮我解决这个问题

Messeges打印出来：

没有返回值或已选择：

Answer 1

您的初始声明仅设置select表中的TimeLog数据，其中明确包含数据。因为那时inner join从这里到其他表，如果那些其他表没有数据，则不会返回任何内容。

要么确保您的Employee，Project和Client表格中包含数据，要么将join更改为left而不是{{1} }}：

inner